Prove that symplectic Lie algebras, sp(n)\mathfrak{sp}(n), are simple

The symplectic lie algebra defined by sp(n)={Xgl2n|XtJ+JX=0} when J=(0II0). So Xsp(n) is of the sort X=(ABCAt) when B=Bt,C=Ct.

This far I was able to get, but how can I prove that it is simple?



Let g=sp(n).

Short answer, from an advanced viewpoint : g is semisimple (because K(x,y)=tr(xy) defines a nondegenerate invariant bilinear form on g), so it will be simple iff its Dynkin diagram is connected. But we know that the diagram is Cn and this is clearly connected.

Longer answer, from a more pedestrian viewpoint : a standard basis B for g is {hi,ai,bi}1in{ci,j,di,j,fi,j,gi,j}1i<jn where

hi=eiien+i,n+i,   ai=ei,i+n,   bi=ei+n,i,

ci,j=eijei+n,j+n,   di,j=ejiej+n,i+n,

fi,j=ei,j+nej,i+n,   gi,j=ei+n,jej+n,i

(and eij is the matrix all of whose coefficients are zero except for the (i,j)-th which is 1).

Now let i be an ideal of g. Then i is stable by ad(x) for any xg. In particular, if H is the vector space spanned by
h1,h2,,hn, then i is stable by all the ad(h), for
hH. Now those ad(h) are simultaneously diagonalized by the basis we have just described above. We deduce that i=HVect(F), where H is a linear subspace of H and F is a subset of B=B{h1,h2,,hn}. Let us now consider the root spaces
ABi=Vect(hi,ai,bi), CDij=Vect(hihj,cij,dij), FGij=Vect(hi+hj,fij,gij)
Each root space R is a Lie subalgebra of g isomorphic to sl2. It easily follows that either iR={0} or
Ri for any R. Combining this with the decomposition of i above, we see that i is the direct sum of some root spaces, plus some subspace of H.

Suppose F contains ai for some i. Then i contains the whole root space Vect(hi,ai,bi). For j>i, i also contains
[ai,gij]=cij, hi[cij,dij]=hj,[hj,aj2]=aj. Similarly, for k<i,
i also contains [ai,gki]=dki, hi+[cki,dki]=hk,[hk,ak2]=ak. So i contains all the ai, and we easily deduce that i=A.

Suppose F contains cij for some i<j. Then i contains the whole root space Vect(hihj,cij,dij). And i also contains hi+hj=[cij,fij]. So i contains both hi and hj, it also contains the root space
Vect(hi,ai,bi), so i=A by the preceding paragraph.

Suppose F contains fij for some i<j. Then i contains the whole root space Vect(hi+hj,fij,gij). And i also contains cij=[fij,bj], so i=A by the preceding paragraph.

Finally we see that i=A unless F is empty. In that case, i is a linear subspace of H. For any i, ad(ai) must be zero on i, otherwise
i would contain ai. It is easily deduced that i={0},
so g is indeed simple.

Source : Link , Question Author : IBS , Answer Author : Ewan Delanoy

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